[Single-digit Integer] Iridium was plated from a solution containing IrCl_y^x- for 2.0 h with a curr — Electrochemistry Chemistry Question

💡 Solution & Explanation

\textbf{Step 1: Calculate total charge and moles of electrons.} \[ Q = I \times t = 0.075 \times 7200 = 540\ \text{C} \] \[ n_e = \frac{Q}{F} = \frac{540}{96500} = 5.596 \times 10^{-3}\ \text{mol} \] \textbf{Step 2: Calculate moles of Ir deposited.} \[ n(\text{Ir}) = \frac{m}{M} = \frac{0.36}{192} = 1.875 \times 10^{-3}\ \text{mol} \] \textbf{Step 3: Find the oxidation state z of Ir.} \[ z = \frac{n_e}{n(\text{Ir})} = \frac{5.596 \times 10^{-3}}{1.875 \times 10^{-3}} \approx 2.98 \approx 3 \] So the oxidation state $z = 3$. \textbf{Step 4: Determine y (number of Cl ligands).} The complex is $[\text{IrCl}_y]^{x-}$. Charge balance: $3 - y = -x$. The common Ir(III) complex with chloride is $[\text{IrCl}_6]^{3-}$: $y = 6$, $x = 3$. \textbf{Step 5: Calculate z + y.} \[ z + y = 3 + 6 = \boxed{9} \]